Answer:
The length of her second displacement = 247.12 m.
The direction of her second displacement = 31.24° from west.
Step-by-step explanation:
As per the question,
From the figure as drawn below,
Let the starting point be O. After running 140 m due west, she reached at point A.
∴ OA = 140 m
And At the end of the run, she is 374 m away from the starting point at an angle of 20° north of west.
∴ OP = 374 m
We have to find the distance AP = x.
By using the cosine rule in triangle OAP
![cos \theta = \frac{OA^{2}+OP^{2}-AP^{2}}{2\times OA\times OP}](https://tex.z-dn.net/?f=cos%20%5Ctheta%20%3D%20%5Cfrac%7BOA%5E%7B2%7D%2BOP%5E%7B2%7D-AP%5E%7B2%7D%7D%7B2%5Ctimes%20OA%5Ctimes%20OP%7D)
After putting the given value, we get
![cos 20= \frac{140^{2}+374^{2}-x^{2}}{2\times 140\times 374}](https://tex.z-dn.net/?f=cos%2020%3D%20%5Cfrac%7B140%5E%7B2%7D%2B374%5E%7B2%7D-x%5E%7B2%7D%7D%7B2%5Ctimes%20140%5Ctimes%20374%7D)
![x^{2}=140^{2}+374^{2} - 2\times 140\times 374\times cos 20](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D140%5E%7B2%7D%2B374%5E%7B2%7D%20-%202%5Ctimes%20140%5Ctimes%20374%5Ctimes%20cos%2020)
∴ x = 247.12 m
Hence,the length of her second displacement = 247.12 m.
Again,
By using the cosine rule in triangle OAP, we get
![cos \alpha = \frac{OA^{2}+AP^{2}-OP^{2}}{2\times OA\times AP}](https://tex.z-dn.net/?f=cos%20%5Calpha%20%3D%20%5Cfrac%7BOA%5E%7B2%7D%2BAP%5E%7B2%7D-OP%5E%7B2%7D%7D%7B2%5Ctimes%20OA%5Ctimes%20AP%7D)
After putting the given value, we get
![cos \alpha = \frac{140^{2}+247.12^{2}-374^{2}}{2\times 140\times 247.12}](https://tex.z-dn.net/?f=cos%20%5Calpha%20%3D%20%5Cfrac%7B140%5E%7B2%7D%2B247.12%5E%7B2%7D-374%5E%7B2%7D%7D%7B2%5Ctimes%20140%5Ctimes%20247.12%7D)
∴ α = 148.759°
Hence, the direction of her second displacement = 180° - α = 180° - 148.759 = 31.24° from west.